Design of Shallow Foundation (Isolated Footings)

Target

Design a square base for a column dimension of 400x400.
Method

Step 1) Given data

Post size 400mmx400mm
Compressive axial load for load combination (1.5 DL + 1.5 LL) is 2000 kN Safe bearing capacity of soil at a depth of 2 meters below E.G.L is 150 kN/m²
Assuming the self-weight of the base plate and the soil above it as a load of 10%.
Concrete class = 25 kN/m²
Assuming fy = 415 N/mm²
Here a sloping base is designed.

Step 2) Base size

Unfactored load 2000/1.5 = 1333.33KN
Safe bearing capacity of soil = 150
Foundation area 1333.55 X 1.1/ 150 = 9.78 m2

Since this is a square root, B = √9.78
B-L = 3.5 m
Foundation area = 12.25 m²
Therefore, the footing area provided is greater than the footing area required 12.25 > 9.78...................Good

Step 3) Net upward soil pressure.

Notupward Pressure ='(factored load)/(specified base area
= (2000)/12.25")

= 163.26 kn/m2

Step 4) Depth of foundations

Maximum bending moment at the column face
Length of the critical section = 3.5-0.2 = 1.55m
Load on critical section = Net upward pressure X Length
= 163.26 x 3.5

= 571.76 kN/m

Moment = 672 X (155) -= 683.4 kN/m
Ultimate moment, Mu = 1.5 X 686.4-1030 kN/m
Depth of foundation = D-Mu -0.138 X fek XBXd
= 1030 x 10 = 0.138 x 0.25 x 3500 x d

= d=√85300 = 292 mm

Using a ruler with 2.5 times the value to check the slip
d = 292 x 2.5 = 730 mm
provides 50mm cover, d = 730+50 780mm-800mm

Step 5) Gross load capacity

Assuming that the unit weight of the soil is 18 kN/m³
Foot load = 1333.3 kN
Pad weight. g = 3.5 x 3.5 x 25 x 0.8 = 245 kN
weight of soil = 3.5 X 3.5 X (2-0.8) X 18 = 261.6 kN
Total load =1852kN
Gross load capacity = 1852)/(3.5x3.5) = 257kN/m²
Since 257 kN/m²>150 kN/m², we need to increase the base size to 3.8m x 3.8m
New total load = 1934KN
gross load capacity 133.9 kN/m²<150kN/m², therefore safe
Corrected net soil pressure

Stop 6) Reinforcement

Percent reinforcement Ast/Bd x 100
P = 3640 x 100/3800 x 800 = 0.119%
Number of rods required for reinforcement = Ast/ Area of ​​one rod (16 mm)
Number of stripes = 3640/(3.14x16x16/4)
Distance between bars = B / Number of bars
Let's say 18 the number of bars with a diameter of 16 mm at 160 mm c/c

Step 7) Check for unidirectional slip

The critical section for shear is at a distance d from the face of the column.
Shear Force (Vu = Net Soil Pressure X B (1/2(B-b)-d)
= 138.5 x 3.8 x (1/2 x (3.8 – 0.4)

= 473.67 kN

Step 8) Check the two-way skid

The critical section for shear is at a distance d/2 from the face of the column
Shear Force, Shear Force (1) = Net Soil Pressure X(B2 – b02)
= 138.5x(3.82-(0.8+0.4)2)

= 1800.5 kN

perimeter of section B0 = 2(1.2+1.2)=4.8m

Vu/ B0d = (1800.5x103)/(4800x800) = 0.46 N/mm²

Permissible shear stress = ks,XTc, (from IS 456 2000)
Pc = 1 (max.)

Tv = 0.25 X √fck

= 0.25 x √25

= 1.25 N/mm2

Tv < Tc. Therefore Ok

Step 9) Length of development

d = 0.87xfs x φ / 4x τbd

= 0.87x25x16/4x1.6x1.4

= 677

Length available

= 1600-50

= 1550 mm > 677 mm

Result

The sloped isolated footing is designed for square Column of size 400mmx400mm
The footing is safe against the bearing pressure